# How do you write the partial fraction decomposition of the rational expression (10x + 2)/ (x^3 - 5x^2 + x - 5)?

##### 1 Answer
Jan 18, 2016

$\frac{10 x + 2}{{x}^{3} - 5 {x}^{2} + x - 5} = \frac{- 2 x}{{x}^{2} + 1} + \frac{2}{x - 5}$

$= - \frac{1}{x - i} - \frac{1}{x + i} + \frac{2}{x - 5}$

#### Explanation:

First factor the denominator by grouping:

${x}^{3} - 5 {x}^{2} + x - 5$

$= \left({x}^{3} - 5 {x}^{2}\right) + \left(x - 5\right)$

$= {x}^{2} \left(x - 5\right) + 1 \left(x - 5\right)$

$= \left({x}^{2} + 1\right) \left(x - 5\right)$

If we stick with Real coefficients for now, then these factors will be the denominators of the partial fraction decomposition.

So we want to solve:

$\frac{10 x + 2}{{x}^{3} - 5 {x}^{2} + x - 5} = \frac{A x + B}{{x}^{2} + 1} + \frac{C}{x - 5}$

$= \frac{\left(A x + B\right) \left(x - 5\right) + C \left({x}^{2} + 1\right)}{{x}^{3} - 5 {x}^{2} + x - 5}$

$= \frac{\left(A + C\right) {x}^{2} + B x + \left(C - 5 B\right)}{{x}^{3} - 5 {x}^{2} + x - 5}$

Equating coefficients of ${x}^{2}$, $x$ and the constant term we find:

(i) $A + C = 0$

(ii) $B - 5 A = 10$

(iii) $C - 5 B = 2$

Combining these using the recipe (i) - (iii) - 5(ii) we find:

$A + C - C + 5 B - 5 B + 25 A = 0 - 2 - 50 = - 52$

So: $A = - \frac{52}{26} = - 2$

Hence: $C = 2$

and $B = 10 + 5 A = 10 - 10 = 0$

So:

$\frac{10 x + 2}{{x}^{3} - 5 {x}^{2} + x - 5} = \frac{- 2 x}{{x}^{2} + 1} + \frac{2}{x - 5}$

Next, if we allow Complex coefficients, then we can factor $\left({x}^{2} + 1\right)$ as $\left(x - i\right) \left(x + i\right)$, hence:

$\frac{- 2 x}{{x}^{2} + 1} = - \frac{1}{x - i} - \frac{1}{x + i}$